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To arrive at its destination on time, a bus should have main

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To arrive at its destination on time, a bus should have main  [#permalink]

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New post 22 Nov 2010, 10:40
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To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

(1) \(V = 60\)
(2) \(X = 20\)

(C) 2008 GMAT Club - m10#8

Thanks!!!
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New post 22 Nov 2010, 18:59
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Knesl wrote:
hi,

can somebody help me with following question? I would need the explanation.
To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

1. \(V = 60\)
2. \(X = 20\)

(C) 2008 GMAT Club - m10#8

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

Thanks!!!


A trick used by GMAT is that they give you similar looking data in the two statements (e.g. in this question, value of one variable given in each statement) and you feel that either each alone will be sufficient or both together. But analyzing the data shows that only one statement is sufficient, the other is not.

The bus covers 1/3 of distance at speed V. The rest 2/3 at speed 1.2V

___________________________
.....V........ ----------1.2V----------

Usual time for the journey is, lets say, T hours, but it arrived in T - X/60 hours. (X is in minutes so you have to change it to hours)

Now, out of the usual T hours, in first 1/3rd of the journey, the bus takes T/3 hours. In rest 2/3rd of the journey, the bus usually takes 2T/3 hours.
During this particular journey, the bus took T/3 hours for the first 1/3rd of the trip (since it traveled at speed V) but for the rest 2/3rd of the trip, it took less time since it traveled at greater speed of 6V/5 (Convert decimals to fractions)
If speed becomes 6/5 of original, time taken will become 5/6 of original.
So time taken to travel 2/3rd of the trip is 5/6 * 2T/3 = 5T/9 hours

In this trip, the total time taken = T/3 + 5T/9 = 8T/9
So the X/60 hours that are saved are equal to T - 8T/9 = T/9

X/60 = T/9

Statement 1: Just V doesn't help. You need D too to get the value of T. Not sufficient.

Statement 2: Knowing just X is sufficient to get T as seen above. Sufficient.

Answer (B).
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Re: Question 8 in m10  [#permalink]

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New post 21 Jan 2011, 10:26
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good question.
there's a difference of X "mins" by varying speed in the last \(\frac{2}{3}\) distance, so from the given info, we can write.

\(\frac{2D}{3V} - \frac{2D}{3*1.2V} = \frac{X}{60}\)

\(\frac{2D}{3V}(1 - \frac{1}{1.2}) = \frac{X}{60}\)

\(\frac{D}{V}= \frac{3X}{20}\)

Note: \(\frac{D}{V} =\) "actual duration" of the trip (and this is what we want to find).

1) we still don't know the value of X, so we can't calculate \(\frac{D}{V}\)

2) \(\frac{D}{V}= \frac{3*20}{20}\)

actual time = 3 hrs .... B is SUFFICIENT
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Re: Question 8 in m10  [#permalink]

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New post 22 Nov 2010, 10:50
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D/v - D/1.2v = X


1.Given V=60 A is not sufficient since we have 2 variables to be resolved in the above equation
2.Given X=20 B is sufficient as-
D/v-D/1.2v = 20
Solving above equation, V is cancelled out leaving value for D which is sufficient.


Answer:- B
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Re: Question 8 in m10  [#permalink]

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New post 22 Nov 2010, 10:59
Thank you Sarang. your answer helped me
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Re: Question 8 in m10  [#permalink]

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New post 23 Nov 2010, 11:06
Karishma,
Awesome explanation
Thanks
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Re: Question 8 in m10  [#permalink]

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New post 14 Mar 2011, 00:51
D - Distance.

D/V - D/3*1/V - 2D/3 * 1/(1.2V) = X

=> D/V(1 - 1/3 - 2/3 * 1/1.2) = X

From (1), V won't cancel out on it's own, so the answer is B.
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Re: To arrive at its destination on time, a bus should have main  [#permalink]

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New post 25 Jan 2014, 10:48
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m10 q08

To arrive at its destination on time the bus should have maintained a speed of \(v\) kilometers per hour throughout the journey. Instead, after going the first third of the distance at \(v\) kilometers per hour, the bus increased its speed and went the rest of the distance at \(1.2v\) kilometers per hour. As a result, the bus arrived at its destination \(x\) minutes earlier than planned. What was the actual duration of the trip?

A bus covered 1/3 of the distance at \(v\) kilometers per hour and the remaining 2/3 of the distance at \(1.2v\) kilometers per hour.

Say the actual duration of the trip is \(t\) and the distance is \(d\);

Then \(t=\frac{(\frac{d}{3})}{v}+\frac{(\frac{d2}{3})}{1.2v}\) --> \(t=\frac{d}{v}*(\frac{1}{3}+\frac{2}{3.6})\) --> \(t=\frac{d}{v}*\frac{8}{9}\)

Also we know that if the speed throughout the journey had been \(v\) kilometers per hour the bus would need \(\frac{x}{60}\) hours more time to cover the same distance: \(t+\frac{x}{60}=\frac{d}{v}\);

Substitute \(\frac{d}{v}\) in the first equation: \(t=(t+\frac{x}{60})*\frac{8}{9}\). So, to get \(t\) we need to know the value of \(x\).

(1) v = 60. Not sufficient.
(2) x = 20. Sufficient.

Answer: B.
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New post 04 Aug 2015, 06:53
VeritasPrepKarishma wrote:
Knesl wrote:
hi,

can somebody help me with following question? I would need the explanation.
To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

1. \(V = 60\)
2. \(X = 20\)

(C) 2008 GMAT Club - m10#8

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

Thanks!!!


A trick used by GMAT is that they give you similar looking data in the two statements (e.g. in this question, value of one variable given in each statement) and you feel that either each alone will be sufficient or both together. But analyzing the data shows that only one statement is sufficient, the other is not.

The bus covers 1/3 of distance at speed V. The rest 2/3 at speed 1.2V

___________________________
.....V........ ----------1.2V----------

Usual time for the journey is, lets say, T hours, but it arrived in T - X/60 hours. (X is in minutes so you have to change it to hours)

Now, out of the usual T hours, in first 1/3rd of the journey, the bus takes T/3 hours. In rest 2/3rd of the journey, the bus usually takes 2T/3 hours.
During this particular journey, the bus took T/3 hours for the first 1/3rd of the trip (since it traveled at speed V) but for the rest 2/3rd of the trip, it took less time since it traveled at greater speed of 6V/5 (Convert decimals to fractions)
If speed becomes 6/5 of original, time taken will become 5/6 of original.
So time taken to travel 2/3rd of the trip is 5/6 * 2T/3 = 5T/9 hours

In this trip, the total time taken = T/3 + 5T/9 = 8T/9
So the X/60 hours that are saved are equal to T - 8T/9 = T/9

X/60 = T/9

Statement 1: Just V doesn't help. You need D too to get the value of T. Not sufficient.

Statement 2: Knowing just X is sufficient to get T as seen above. Sufficient.

Answer (B).


I almost got to the last step but could not arrive at the answer...can someone help..Pls...

my steps:

let distance = d
new avg speed = \(\frac{total dist}{total time}\) = \(\frac{3.4vd}{t-x}\)

equating the 2 distances

\(v*t = 3.4 V*d\)

t=3.4d

I got stuck here...
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New post 04 Aug 2015, 09:23
robinpallickal wrote:

I almost got to the last step but could not arrive at the answer...can someone help..Pls...

my steps:

let distance = d
new avg speed = \(\frac{total dist}{total time}\) = \(\frac{3.4vd}{t-x}\)

equating the 2 distances

\(v*t = 3.4 V*d\)

t=3.4d

I got stuck here...


I do not understand the expression 3.4vd/(t - x) and also how you arrived at it. Also, x is in minutes while speed is in miles per hour so t should be in hours. Hence t - x doesn't mean anything. It must be t - x/60.
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To arrive at its destination on time, a bus should have main  [#permalink]

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New post Updated on: 05 Aug 2015, 19:26
VeritasPrepKarishma wrote:
robinpallickal wrote:

I almost got to the last step but could not arrive at the answer...can someone help..Pls...

my steps:

let distance = d
new avg speed = \(\frac{total dist}{total time}\) = \(\frac{3.4vd}{t-x}\)

equating the 2 distances

\(v*t = 3.4 V*d\)

t=3.4d

I got stuck here...


I do not understand the expression 3.4vd/(t - x) and also how you arrived at it. Also, x is in minutes while speed is in miles per hour so t should be in hours. Hence t - x doesn't mean anything. It must be t - x/60.


I completely misread the question. My bad. Also I should have taken x in mins.

Thanks Karishma.

Originally posted by robinpallickal on 05 Aug 2015, 19:03.
Last edited by robinpallickal on 05 Aug 2015, 19:26, edited 1 time in total.
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Re: To arrive at its destination on time, a bus should have main  [#permalink]

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New post 05 Aug 2015, 19:24
robinpallickal wrote:
VeritasPrepKarishma wrote:
robinpallickal wrote:

I almost got to the last step but could not arrive at the answer...can someone help..Pls...

my steps:

let distance = d
new avg speed = \(\frac{total dist}{total time}\) = \(\frac{3.4vd}{t-x}\)

equating the 2 distances

\(v*t = 3.4 V*d\)

t=3.4d

I got stuck here...


I do not understand the expression 3.4vd/(t - x) and also how you arrived at it. Also, x is in minutes while speed is in miles per hour so t should be in hours. Hence t - x doesn't mean anything. It must be t - x/60.


Avg speed = \((\frac{v*d}{3} + \frac{1.2*d*2}{3})/(t-x)\)
= \(\frac{(v*d+2.4vd)}{(3(t-x))}\)

I think my mistakes are I did not carry forward 3 in the subsequent equations and did not convert x to minutes. Can you Pls. confirm.


Yes you are correct. In quant question rated to distance-speed or work-rate problems, always make sure that the rates, times or distances are cocnsistent throughout. Any deviation int he units will lead to incorrect answers.
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New post 25 Aug 2015, 10:20
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I solved the question as below :

Distance = D + D + D ---------------------------- (Total distance = 3D)
Speed = V , 1.2V , 1.2V -----------------------------(Speed for each part)
Time = T + (5/6)T + (5/6)T -----------------------------(Time taken for each part)
Reduced Time = (1/6)T + (1/6)T

Using option B we know that the reduced time is 20 = (1/6T)*2
Hence B.
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New post 25 Aug 2015, 21:23
NickHalden wrote:
I solved the question as below :

Distance = D + D + D ---------------------------- (Total distance = 3D)
Speed = V , 1.2V , 1.2V -----------------------------(Speed for each part)
Time = T + (5/6)T + (5/6)T -----------------------------(Time taken for each part)
Reduced Time = (1/6)T + (1/6)T

Using option B we know that the reduced time is 20 = (1/6T)*2
Hence B.


Great use of Speed-Time ratios!

Just note that it should be
20/60 = (1/6)2T

because 20 is in mins but everything else is in hours.
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New post 26 Nov 2015, 16:02
Question can be reformulated as: can you find the value of d/v?

T1 = d/v
T2 = 1/3*d/v + 5/6*2/3*d/v
(1/3 + 5/9)d/v
d/v*8/9
T2 = T1 - x/60

8/9*d/v = d/v - x/60

d/v - 8/9*d/v = x/60

1/9.d/v = x/60
d/v = 9*.x/60

From this we can conclude that the value of d/v can be obtained by knowing x or by knowing d&v together. Therefore statement 1 is sufficient and statement 2 is insufficient. Answer is A.
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Re: To arrive at its destination on time, a bus should have main  [#permalink]

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New post 29 Nov 2015, 11:12
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Knesl wrote:
To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

(1) \(V = 60\)
(2) \(X = 20\)

(C) 2008 GMAT Club - m10#8

Thanks!!!


--------------------------------------------------------------------------------------
Always remember, in case of inverse relationship such as this an increase of 1/x times of V (velocity) should be accompanied by a decrease of 1/(x+1) in T (time), so as to keep the overall value = V*T = distance constant.

There V -> 1.2 V hence V increases by 20% or 1/5 times, hence the time should be decrease by 1/6th time.

Now, it's already given that the 1/6th of T is X ( in other words the change in duration is already given where the T is the actual duration time)

thus (1/6) * T =X ; T=6X.

Hence just knowing X will give you the solution. So, here B itself is enough, but A is not enough. Answer - B.
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New post 27 Jan 2019, 23:02
Knesl wrote:
To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

(1) \(V = 60\)
(2) \(X = 20\)

(C) 2008 GMAT Club - m10#8

Thanks!!!


@vertiaskarishma Bunuel chetan2u

With alone A is also correct
Since 1/3 of the distance is 60 km and it took one our to cover it
Since bus should have maintained same speed i.e 60 km/h so it will take additional 2 hr to cover the distance so the actual trip time will be 3 hrs
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Re: To arrive at its destination on time, a bus should have main  [#permalink]

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New post 27 Jan 2019, 23:18
teaserbae wrote:
Knesl wrote:
To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

(1) \(V = 60\)
(2) \(X = 20\)

(C) 2008 GMAT Club - m10#8

Thanks!!!


@vertiaskarishma Bunuel chetan2u

With alone A is also correct
Since 1/3 of the distance is 60 km and it took one our to cover it
Since bus should have maintained same speed i.e 60 km/h so it will take additional 2 hr to cover the distance so the actual trip time will be 3 hrs


Hey teaserbae!

You have assumed a total distance of 180 km (speed of 60 kmph for 1 hr covering a distance of 60 km).
All we know is that 1/3rd the distance is at a speed of 60 kmph. The total distance could very well be 300 km such that 100 km was covered at 60 kmph in a total time of 100/60 = 5/3 hrs
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To arrive at its destination on time, a bus should have main  [#permalink]

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New post 17 Aug 2019, 06:53
Bunuel VeritasKarishma I have a small doubt. I understand the solutions provided. However, I tried to do it using a slightly different but my final equation is different. Kindly help.
I assumed that since the train travelling with 0.2V speed across 2/3D caused it to reach early by X/60 min.
So,

X/60= {(2/3)*D/(4/5)*V}
This is leading to (5/6)*(D/V)=X/60.

But as per other solutions the equation is T/9=X/60. Can you please help me understand where am I going wrong?
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